The volume of a cube is increasing at a rate of $18$ cubic meters per hour. At a certain instant, the volume is $8$ cubic meters. What is the rate of change of the surface area of the cube at that instant (in square meters per hour)? Choose 1 answer: Choose 1 answer: (Choice A) A $36$ (Choice B) B $24$ (Choice C) C $\dfrac{3}{2}$ (Choice D) D $(\sqrt[3]{18})^{^2}$
Solution: Setting up the math Let... $s(t)$ denote the cube's side at time $t$, $V(t)$ denote the cube's volume at time $t$, and $S(t)$ denote the cube's surface area at time $t$. We are given that $V'(t)=18$, We are also given that $V(t_0)=8$ for a specific time $t_0$. We want to find $S'(t_0)$. Relating the measures $S(t)$ and $s(t)$ relate to each other through the formula for the surface area of a cube: $S(t)=6[s(t)]^2$ We can differentiate both sides to find an expression for $S'(t)$ : $S'(t)=12s(t)s'(t)$ $V(t)$ and $s(t)$ relate to each other through the formula for the volume of a cube: $V(t)=[s(t)]^3$ We can differentiate both sides to find an expression for $V'(t)$ : $V'(t)=3[s(t)]^2s'(t)$ Using the information to solve Let's plug ${V(t_0)}={8}$ into the expression for $V(t_0)$ : $\begin{aligned} {V(t_0)}&=[s(t_0)]^3 \\\\ {8}&=[s(t_0)]^3 \\\\ {2}&={s(t_0)} \end{aligned}$ Let's plug ${V'(t_0)}={18}$ and ${s(t_0)}={2}$ into the expression for $V'(t_0)$ : $\begin{aligned} {V'(t_0)}&=3[{s(t_0)}]^2s'(t_0) \\\\ {18}&=3({2})^2s'(t_0) \\\\ C{\dfrac{3}{2}}&=C{s'(t_0)} \end{aligned}$ Now let's plug ${s(t_0)}={2}$ and $C{s'(t_0)}=C{\dfrac{3}{2}}$ into the expression for $S'(t_0)$ : $\begin{aligned} S'(t_0)&=12{s(t_0)}C{s'(t_0)} \\\\ &=12\left({2}\right)\left(C{\dfrac{3}{2}}\right) \\\\ &=36 \end{aligned}$ In conclusion, the rate of change of the surface area of the cube at that instant is $36$ square meters per hour. Since the rate of change is positive, we know that the surface area is increasing.